# What is the outer product of a tensor product of vectors?

• I
If one has two single-particle Hilbert spaces ##\mathcal{H}_{1}## and ##\mathcal{H}_{2}##, such that their tensor product ##\mathcal{H}_{1}\otimes\mathcal{H}_{2}## yields a two-particle Hilbert space in which the state vectors are defined as $$\lvert\psi ,\phi\rangle =\lvert\psi\rangle\otimes\lvert\phi\rangle\in\mathcal{H}_{1}\otimes\mathcal{H}_{2}$$ where ##\lvert\psi\rangle\in\mathcal{H}_{1}## and ##\lvert\phi\rangle\in\mathcal{H}_{2}##.

Now, the inner product for ##\mathcal{H}_{1}\otimes\mathcal{H}_{2}## is defined such that $$\langle\phi ,\psi\vert\psi ,\phi\rangle =\left(\langle\phi\rvert\otimes\langle\psi\lvert\right)\left(\lvert\psi\rangle\otimes\lvert\phi\rangle\right) =\langle\psi\lvert\psi\rangle_{1}\langle\phi\lvert\phi\rangle_{2}$$ where ##\langle\cdot\lvert\cdot\rangle_{1}## is the inner product defined on ##\mathcal{H}_{1}## and ##\langle\cdot\lvert\cdot\rangle_{2}## the inner product defined on ##\mathcal{H}_{2}##.

How though is the outer product defined? Is it simply $$\lvert\psi ,\phi\rangle\langle\phi ,\psi\rvert =\lvert\psi\rangle\otimes\lvert\phi\rangle\langle\phi\rvert\otimes\langle\psi\rvert =\lvert\psi\rangle\langle\psi\rvert_{1} \lvert\phi\rangle\langle\phi\rvert_{2}$$ where ##\lvert\psi\rangle\langle\psi\rvert_{1}## is the outer product in ##\mathcal{H}_{1}## and ##\lvert\phi\rangle\langle\phi\rvert_{2}## is the outer product in ##\mathcal{H}_{2}##.

andrewkirk
Homework Helper
Gold Member
It might be easier to think of more general outer products, rather than the outer product of an element with itself, as you have written.
That is, consider the outer product ##|a,b\rangle\langle c,d|=|a\rangle\otimes |b\rangle\langle c|\otimes\langle d|##.

This is an operator on the product space ##\mathcal{H}_{1}\otimes\mathcal{H}_{2}##. To see what it does, we apply it to an element of that space ##|e,f\rangle=|e\rangle\otimes |f\rangle##.

The result is ##|a,b\rangle \langle c|e\rangle \langle d|f\rangle##

This is an operator on the product space H1⊗H2\mathcal{H}_{1}\otimes\mathcal{H}_{2}. To see what it does, we apply it to an element of that space |e,f⟩=|e⟩⊗|f⟩|e,f\rangle=|e\rangle\otimes |f\rangle.

The result is |a,b⟩⟨c|e⟩⟨d|f⟩

So one can't define it as ##\lvert a,b\rangle\langle c,d\rvert =\lvert a\rangle\langle c\rvert\lvert b\rangle\langle d\rvert## then? Should it be something like $$\lvert a,b\rangle\langle c,d\rvert =\lvert a\rangle\langle c\rvert\otimes\lvert b\rangle\langle d\rvert$$

I ask in particular as I'm trying to understand the notion of a partial trace.

Last edited:
andrewkirk
Homework Helper
Gold Member
So one can't define it as ##\lvert a,b\rangle\langle c,d\rvert =\lvert a\rangle\langle c\rvert\lvert b\rangle\langle d\rvert## then?
That symbol string on the RHS is not intrinsically meaningful. It juxtaposes the two operators ##\lvert a\rangle\langle c\rvert## and ##\lvert b\rangle\langle d\rvert## but:
- since they are not scalars, the juxtaposition cannot be interpreted as scalar multiplication
- since they are operators on different spaces (##\mathcal H_1## and ##\mathcal H_2##), the juxtaposition cannot be interpreted as composition of operation.

Should it be something like $$\lvert a,b\rangle\langle c,d\rvert =\lvert a\rangle\langle c\rvert\otimes\lvert b\rangle\langle d\rvert$$
That is one of a number of intrinsically meaningful ways of writing it. It has the same meaning as what I wrote above ##(
|a\rangle\otimes |b\rangle)(\langle c|\otimes\langle d|)## (I have added parentheses here that were only implied above, to make it clear what operations are being performed).

That is one of a number of intrinsically meaningful ways of writing it. It has the same meaning as what I wrote above (|a⟩⊗|b⟩)(⟨c|⊗⟨d|)

How does one show that the two expressions are equivalent? Would it be something like this:
$$\left(\lvert a\rangle\otimes\lvert b\rangle\langle c\rvert\otimes\langle d\rvert\right)\lvert e\rangle\otimes\lvert f\rangle =\lvert a\rangle\otimes\lvert b\rangle\left(\langle c\vert e\rangle\langle d\vert f\rangle\right)\\ =\left(\lvert a\rangle\langle c\vert e\rangle\right)\otimes \left(\lvert b\rangle\langle d\vert f\rangle\right) =\left(\lvert a\rangle\langle c\lvert\otimes\lvert b\rangle\langle d\lvert\right)\lvert e\rangle\otimes\lvert f\rangle$$

andrewkirk
$$(|a,b\rangle\langle c,d|)|e,f\rangle\equiv (|a\rangle\otimes|b\rangle)(\langle c|\otimes\langle d|)(|e\rangle\otimes|f\rangle)=(\langle c|e\rangle|a\rangle)\otimes (\langle d|f\rangle|b\rangle)$$
$$(|a\rangle\langle c|\otimes|b\rangle\langle d|)|e,f\rangle\equiv (|a\rangle\langle c|\otimes|b\rangle\langle d|)(|e\rangle\otimes |f\rangle)\ =(|a\rangle\langle c|)|e\rangle\otimes (|b\rangle\langle d|)|f\rangle \equiv (\langle c|e\rangle|a\rangle)\otimes (\langle d|f\rangle|b\rangle)$$
$$(|a,b\rangle\langle c,d|)=(|a\rangle\langle c|\otimes|b\rangle\langle d|)$$